4
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The state vector is $$ \textbf{X} = \begin{bmatrix} x \\ y \\ v_{x} \\ v_{y} \end{bmatrix}$$

transition function is $$ \textbf{X}_{k} = f(\textbf{X}_{k-1}, \Delta t) = \begin{cases} x_{k-1} + v_{xk} \Delta t \\ y_{k-1} + v_{yk} \Delta t \end{cases} $$

$z_{b} = atan2(y, x)$ and $z_{r} = \sqrt{ x^{2} + y^{2}}$

the Jacobian of the observation model: $$ \frac{\partial h}{\partial x} = \begin{bmatrix} \frac{-y}{x^{2}+y^{2}} & \frac{1}{x(1+(\frac{y}{x})^{2})} & 0 & 0 \\ \frac{x}{\sqrt{ x^{2} + y^{2}}} & \frac{y}{\sqrt{ x^{2} + y^{2}}} & 0 & 0 \end{bmatrix} $$

My question is how the Jacobian of the observation model has been obtained? and why it is 2X4?

the model from Kalman filter.

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closed as unclear what you're asking by Mark Booth Oct 31 '13 at 18:58

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Rather than appending the answer to the question, add your answer as an answer -- this is both allowed and encouraged, since it will indicate that the question has an answer. $\endgroup$ – Ian Oct 31 '13 at 4:04
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The Jacobian is of size $2\times 4$ because you have four state elements and two measurement equations.

The Jacobian of the measurement model is the matrix where each $i,j$ element corresponds to the partial derivative of the $i$th measurement equation with respect to the $j$th state element.

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  • 1
    $\begingroup$ thanks for being informative and helpful. This is what I thought. $\endgroup$ – CroCo Nov 1 '13 at 15:08
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I've solved the problem by applying $$ \frac{\partial h}{\partial x} = \begin{bmatrix} \frac{\partial z_{b}}{\partial x} & \frac{\partial z_{b}}{\partial y} & \frac{\partial z_{b}}{\partial v_{x}} & \frac{\partial z_{b}}{\partial v_{y}} \\ \frac{\partial z_{r}}{\partial x} & \frac{\partial z_{r}}{\partial y} & \frac{\partial z_{r}}{\partial v_{x}} & \frac{\partial z_{r}}{\partial v_{y}} \end{bmatrix} = \begin{bmatrix} \frac{-y}{x^{2}+y^{2}} & \frac{1}{x(1+(\frac{y}{x})^{2})} & 0 & 0 \\ \frac{x}{\sqrt{x^{2}+y^{2}}} & \frac{y}{\sqrt{x^{2}+y^{2}}} & 0 & 0 \\ \end{bmatrix} $$

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  • $\begingroup$ This is not a correct answer to the question, or the question was not worded correctly. Either way it is misleading for future readers. $\endgroup$ – Josh Vander Hook Oct 31 '13 at 17:14
  • $\begingroup$ @CroCo - Please update either your question or answer. As far as I can see, your question isn't very clear and this answer doesn't answer what you appear to have asked. $\endgroup$ – Mark Booth Oct 31 '13 at 18:58
  • $\begingroup$ @MarkBooth, thanks for the advice. Please remove the put on hold. $\endgroup$ – CroCo Nov 1 '13 at 15:10
  • $\begingroup$ Sorry @CroCo, but I still don't see how your answer is related to the actual question that you asked, which was My question is how the Jacobian of the observation model has been obtained? and why it is 2X4? $\endgroup$ – Mark Booth Nov 2 '13 at 0:17
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    $\begingroup$ I don't see any problem with my answer except that the element (1,2) seems wrong in the given model. It should be $$\frac{x}{y^{2+x^{2}}}$$. That's all I can say. If you think this is a wrong answer, would you kindly suggest the correct one instead of putting it on hold. It is easy to criticize a guy, but it is not easy to guide him/her. It is up to you, I don't really care any more. Any way, thanks again. $\endgroup$ – CroCo Nov 2 '13 at 5:55

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